CAT 2024 Slot 2 QA Question 21

Type-in-the-answer (no negative marking) · Algebra · Polynomials · Try it, then check the answer and solution below.

CAT 2024 Slot 2QAAlgebra • PolynomialsModerate

x

and

y

are real numbers such that

4x^2 + 4y^2 - 4xy - 6y + 3 = 0 , then the value of

(4x + 5y)$ is

TITA Answer:

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Official Convenor Answer Key: 7

Answer and solution

Answer: 7

📌 Core Concept

To solve the given quadratic equation in two variables, we can treat it as a quadratic in one variable and use the discriminant condition for real solutions.

🔢 Step-by-Step Solution

Rewrite the equation:

4x^2 + 4y^2 - 4xy - 6y + 3 = 0 Treat it as a quadratic in x: 4x^2 - 4xy + (4y^2 - 6y + 3) = 0

Identify coefficients:

a = 4

b = -4y

c = 4y^2 - 6y + 3

Apply the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute the coefficients:

x = \frac{4y \pm \sqrt{16y^2 - 16(4y^2 - 6y + 3)}}{8}

Simplify the discriminant:

\sqrt{16y^2 - 64y^2 + 96y - 48} = \sqrt{-48(y - 1)^2}

For real solutions, the discriminant must be non-negative:

- 48(y - 1)^2 \geq 0 \implies y = 1

Sub

CAT 2024 Slot 2 QA Question 21

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